Question

数列（an}的前n项和为Sn,且满足an=-3Sn*Sn-1(n>=2),a1=1/3 1.证（1/Sn}是等差数列

Answer 1

1）S[n]-S[n-1] = -3S[n]*S[n-1]

两边除以 - S[n]*S[n-1]

1/S[n] -1/S[n-1] =3

2）S[n]=1/(3n)

a[n]= S[n] - S[n-1] =1/( 3(1-n)n )

b[n]=2/(3n)

因为1/(n^2) < 1/( (n-1)n ) =1/(n-1) - 1/n

b2平方+b3平方+.+bn平方

=4/9*(1/2^2 +1/3^2+...+1/n^2)